Question

If value of \[\tan {35^ \circ } = k\] is given then the value of \[\dfrac{{\tan {{145}^ \circ } - \tan {{125}^ \circ }}}{{1 + \tan {{145}^ \circ }\tan {{125}^ \circ }}}\] is
A) \[\dfrac{{2k}}{{1 - {k^2}}}\]
B) \[\dfrac{{2k}}{{1 + {k^2}}}\]
C) \[\dfrac{{1 - {k^2}}}{{2k}}\]
D) \[\dfrac{{1 - {k^2}}}{{1 + {k^2}}}\]

Answer 1

Hint: We must remember that \[\tan (90 - \theta ) = \cot \theta \] and also \[\tan (90 + \theta ) = - \cot \theta \] we will use these and change the following angles in \[90 + \theta \] then let's see where we end up.
Complete Step by Step Solution:
We are given that \[\dfrac{{\tan {{145}^ \circ } - \tan {{125}^ \circ }}}{{1 + \tan {{145}^ \circ }\tan {{125}^ \circ }}}\]
Here we can write \[145 = 90 + 55\& 125 = 90 + 35\]
\[\therefore \tan {145^ \circ } = \tan {(90 + 55)^ \circ }\& \tan {125^ \circ } = \tan {(90 + 35)^ \circ }\]
As we know that \[\tan (90 + \theta ) = - \cot \theta \]
Using this we will get \[\tan {145^ \circ } = - \cot {55^ \circ }\& \tan {125^ \circ } = - \cot {35^ \circ }\]
Now the whole thing will look like
\[\dfrac{{ - \cot {{55}^\circ } - \left( { - \cot {{35}^\circ }} \right)}}{{1 + \left( { - \cot {{55}^\circ }} \right)\left( { - \cot {{35}^\circ }} \right)}}\]
If we try to solve further we will get it as
\[\begin{array}{l}
 \Rightarrow \dfrac{{\cot {{35}^\circ } - \cot {{55}^\circ }}}{{1 + \left( {\cot {{55}^\circ }} \right)\left( {\cot {{35}^\circ }} \right)}}\\
 \Rightarrow \dfrac{{\dfrac{1}{{\tan {{35}^ \circ }}} - \tan {{35}^ \circ }}}{{1 + \dfrac{1}{{\tan {{35}^ \circ }}} \times \tan {{35}^ \circ }}}\\
 \Rightarrow \dfrac{{\dfrac{{1 - {{\tan }^2}{{35}^ \circ }}}{{\tan {{35}^ \circ }}}}}{{1 + 1}}\\
 \Rightarrow \dfrac{{1 - {{\tan }^2}{{35}^ \circ }}}{{2\tan {{35}^ \circ }}}
\end{array}\]
Now as it is given in the question \[\tan {35^ \circ } = k\]
Substituting this we will get it as
\[\dfrac{{1 - {k^2}}}{{2k}}\]
Therefore C is the correct option here

Note: It should be noted that \[\cot \theta = \dfrac{1}{{\tan \theta }}\] which i have used in the solution, you can also try to convert everything in \[\sin \theta \& \cos \theta \] and then solve it, still you will get the same answer but it will be a lot chaotic.